Understanding Electrode Potentials: Why is 1/Molarity Used in the Calculation?

[Puchinita5]

I don't know why, but I just can't figure this out.

"Calculate the electrode potential for a silver filled electrode in a solution of 0.00550 M AgNO3 solution."

I got that

Ag+ + e- --> Ag(s) had an E* = .80V

I got that the answer is E=.80 - .0592 log (1/.00550M)= .6655 V

But why do you put 1/.00550M in the parenthesis?

My first thought was that AgNO3 + e- --> Ag(s) + NO3-
and that the concentreation of AgNO3 is the same as NO3.

so that it would be log(1) , which is wrong.

I just don't get the concept I guess.

 

[Borek]

It all depends on how you write the Nernst equation. There are two conventions:

$$E = E_0 + \frac {RT}{nF} ln \frac{[Ox]}{[Red]}$$

and

$$E = E_0 - \frac {RT}{nF} ln \frac{[Red]}{[Ox]}$$

(personally I prefer the first one, but that's just because I was taught this way).

Note that both equations are equivalent thanks to the sign change.

Oxidized form of silver is Ag⁺, reduced form is a metallic Ag - with activity equal to 1. So you can write the equation either as

$$E = 0.88 + 0.059\ log([Ag^+])$$

or

$$E = 0.88 - 0.059\ log\big(\frac 1 {[Ag^+]}\big)$$

Concentration of Ag⁺ is 0.00550 M, concentration of NO₃^- doesn't matter.

 

[Puchinita5]

oh! okay that makes sense. I don't know why but I always confuse myself with these electrode problems.

Thank you!